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Re: Finding a formula for a sum


In a message dated 6/25/02 5:10:35 AM, Matthias.Bode at oppenheim.de writes:

>when trying to find the number of diagonals in a convex polygon (all corners
>on the perimeter of a circle) with n corners I came across that sum:
>
>+(n-3) ; up to here = diagonals in a triangle.
>
>+(n-3) ; up to here = diagonals in a quadrangle.
>
>+(n-4) ; up to here = diagonals in a pentagon.
>
>+(n-5) ; up to here = diagonals in a hexagon.
>
>+(n-6) ; up to here = diagonals in a heptagon.
>and so on.
>
>I found the general formula (n*n -3*n)/2 with pencil and paper.
>
>How could I coax MATHEMATICA into helping me to find the generalization
>in
>this case - and of course for more difficult ones as well?

It is just the number of ways of taking n things 2 at a time except  
that you must exclude the sides of the polygon.

Binomial[n, 2]-n // Simplify

(1/2)*(-3 + n)*n


Bob Hanlon
Chantilly, VA  USA


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